10 Graph Theory:: Bellman Ford's Algorithm with CSES 10 High Score (1673)

[@sambhavsinha8844]

Your teaching approach reminds me of one of my previous teachers, who was unparalleled (atleast for me) ... Thank you sir for such content

[@leviackerman9882]

I couldn't understand anywhere why it needs to run at most n-1 times And you made it super easy with that example Thank you so much ❤️

[@rahulsawant2093]

Because there can be at most n-1 edges in the graph, so to resolve (relax) each edge length, we should take the worst case scenario.

[@anubhavagrawal3963]

Probably the best videos for graph theory on youtube. hats off.

[@dardev8360]

Thank you. I will be more active henceforth. Any ideas on how we can become more popular?

[@s.d8756]

@@dardev8360 Sir, use a facecam for your videos. It may help. also consistency can help. Create decent social media personality. Lastly, just put out quality content like this, nothing beats content. Once you put out content, it is just waiting for the masses to discover this beautiful channel.

[@satyamchhatrala93]

Very nice playlist !! Highly Appreciated

[@Mr_D0T]

Man you are a gem , i was so confused after reading solution in positive cycles

[@dardev8360]

Man, you are THE gem. You took out the time to write such a sweet comment. Any ideas on how we can become more popular?

[@rounakagarwal1600]

Couldn't find the best video other than this on youtube

[@amoghhuddar6408]

Great work dude! Was so confused on why we require n-1 relaxations to propagate anything and you just explained it all so nicely. Thanks a lot man! Also some newer tc were added on cses like a graph with a single node with self loop with weight 1. You will have to run n relaxations on it cause n-1 is just 0. Do make more videos. Amazing Content❤

[@youknowwho321]

solution starts at 29:00 ( For those who already knows Bellman-Ford's Algorithm)

[@manusisodia5814]

for those whose solution is not accepting add this condition if(n==1) { bool check=false; for(auto e: edges) { if(e.third

[@m.e.t.a.l3125]

You are very much underrated! Awesome channel to learn Graphs..

[@paragroy5359]

Thanks for this playlist really helpful...

[@dardev8360]

Cycle Detection in Directed Graph coming up in a few hours! Stay tuned and Stay updated

[@loudcapricorn5513]

Really very great learning video. 💯 Especially relating everything to past learned topics like Djikstra, BFS, etc

[@dardev8360]

Glad you liked it

[@crazyfootball2271]

The series is good. In this particular video , there is a mistake at 5:21. You have shown 2 graphs , Djikstra will work for the first one but not for the second one. Djisktra would only fail if there is a negative cycle (not a negative edge). Hence , it would work for example at 5:21.

[@abhishekpandey7859]

yes that what i was also thinking

[@yashbansal1520]

It will not work in case in case he is taking a visited array else if we are relaxing the nodes without a visited array it will work.

[@jaishriharivishnu]

seriously! you made it damn easy, I was complicating it.

[@matheuscardoso1110]

this video is really great, congratz!

[@manaskumarpanda3343]

i'm addicted to your videos for cses problemset problems 💚

[@dardev8360]

That's how I want it to be! Stay addicted to the good stuff, man! Looking forward to suggestions.

[@priyankarrajgupta4198]

Very well explained. I dont know why ur videos have such a low view count Are u using Windows XP ?

[@dardev8360]

I also don't know why low view count. I use Windows 7 with Classic Theme.

[@dardev8360]

Share more please? So my views increase and do subscribe!

[@AnkitMishra-sq9nf]

@@dardev8360 sir, make your video titles that people will search for normally, dont put in video number as the first thing in your title , just give simple headings and ue keywords ony so that your channel is shown in search results....like "bellmanford algo" or "cses graph algo qustions"......there are more searches for bell man ford algo rather than csse playlist and even less searches for the big title that you generally put in your titles.....you can put rest of the info in your descriptions...

[@pabitrakb5291]

Best postmortem of Bellman Ford algo ever possible ❤

[@abhipatel13311]

Great content brother🙏🏻

[@sibashissom3404]

Thank you so much for this.

[@nishuz]

Nice video. But at 1:33, the path from 1 to 4 should be 4, not 2, right?

[@dardev8360]

Right. Sorry! :( Thanks for pointing it out.

[@s.d8756]

Extremely Helpful video. Thanks a lot. Are your courses availabel on any other platforms? and why you have only videos on graph? Would love to see your vieos on other topics as well.

[@androidguruji4542]

superb explanation really helpful , thank you . keep uploading new videos love to watch it........

[@rupeshkataria81]

I have one doubt since there are m edges so we should start loop from 1 to m instead of 1 to n in bellman_ford function because your code fail for this input when n=1,m=1 and the edges are 1 1 1 . So edges start from 1 and end at 1 and the weight of edge is 1. its right answer is -1 but your code give 0 .Since for n=1 I our bellman_ford function would not run if I use m instead of n it give wrong output on many test cases

[@haribooshan7457]

I had just changed the number of iterations to max(n-1,1) instead of n-1. This seems to work.

[@mohitahlawat4944]

Dijkstra's will work if there is a negative edge. only thing is that the graph should not have any negative edge cycle

[@anishsinghdhami2557]

It only worked, because of our implementation details. But according to the concept he taught it won't work because we won't be visiting or considering those vertices again which have been processed once(i.e. those vertices which is once taken out of the min_heap) will not be considered to do the relaxation for the remaining vertices. Hope the explanation makes sense if not do a google search and look for some of the stack-overflow articles.

[@sachinsingh1956]

this was a great solution i have ever seen.

[@aadityaupadhyay6264]

Thanks for this video. Wonderful explanation

[@dardev8360]

You are welcome Aaditya. I am impressed that you have kept up with the series! I will be making Floyd-Warshall and posting tonight! Stay tuned!

[@aadityaupadhyay6264]

@@dardev8360 ya sure brother, waiting for more videos to learn the concepts

[@159_mdamirulislam8]

This man deserves more views and likes. Thanks.

[@tarunjayadevan5981]

sir you are amazing ! Thanks so much for such videos .... really helpful ..

[@dardev8360]

I am glad that I am able to help! Will make more content. What else would you like to see?

[@raghavmittal5352]

Great explanation

[@sdmfslkdm]

if we are adding the cap to the dist[v] to inf , wouldn't it make it our answer to change

[@bhanuprasadbellam3107]

Good explanation

[@sommayghosh4617]

I had a doubt your solution is correct and gets accepted , but the idea behind bellman ford is that after n-1 times , if we are again getting a shorter path for the nth time then there is a negative weight cycle but it doesn't get accepted few wrong answers came, but when i did like u , i.e again for n-1 times it got AC, can u tell the intuition behind it?

[@yath3681]

The negative weight cycle that we are getting should also be involved in the 'shortest path' from 1 -> n in our negated graph. If we found a negative cycle in our graph , but it is in no way involved ini the 'shortest path' , then the answer would not be -1.

[@vinayverma8897]

Bro can you share the code pls

[@shishankrawat2105]

"You may go through a tunnel several times." And we don't necessarily have to stop when we reach the Nth room. This means that if we can reach the Nth room with some distance, say d... and then from the Nth room back to room 1, such that now the distance of room 1 is less than 0. So we can re-iterate this path again and again. So, to confirm it what we can do is run cycle N times and then another (N-1) times and then finally see the distance values.

[@dhairya7460]

Sir I have a doubt please help me out. What's my claim is that if we reaches a node that doesn't have any outgoing edges we simply popout that node from priority queue so at the time of poping out if we mark that node as visited then we can't have to face negative cycle. Am i correct sir?

[@KhemendraBhardwaj]

did you able to find it out?

[@Rajjain_]

these are really good. thanks

[@beonthego8725]

Thank you Dardev

[@dardev8360]

You are welcome!

[@kaukuntlaprudviraj108]

Is it mandatory to convert the edge weights to it's negative form, in relaxation why can't we maximize the distance in place of minimizing

[@optimus_prime01]

do a dry run, that's how you learn

[@madhvansharma5917]

Hi, I had a question. I took INF as LLONG_MAX and NINF as -INF. I saw that the code didn't work for some testcases whereas when I took the values of INF and NINF same as in the video, then the Code executed nicely. Could u tell me why is that ?

[@AkashKumar-im2zy]

It is due to overflow,(The maximum value that the variable can store is LLONG_MAX), LLONG_MAX+d will result in overflow, whereas some user-defined infinity like INF= 1e16 will not overflow because INF + u value is still less than the max size the long long variable can store!

[@prime_edits007]

well explained dude :P

[@dardev8360]

Thanks, dude! :P

[@paragsaini2795]

Your r awesome!! Great explanation. 🔥

[@NiteshKumar-xm3nq]

your code is giving wrong answers for self loops . try submitting it .

[@litcode702]

excellent...may I know from where you learnt in this much depth?? nowhere this kind of explanation is given

[@dardev8360]

Glad you liked the video. I learned stuff on my own. Do you want me to continue the series? :D

[@lichuphy]

2 words: THANK YOU

[@dardev8360]

You are welcome! Subscribe for more!

[@gantagourish7536]

I have a doubt. let the input be 5 5 1 2 1 2 3 1 3 4 1 4 2 1 4 5 1 according to your code it will give -1 because 2->3->4->2 will form negative cycle(as edges weight are converted to -ve) but when we draw graph on paper we can see path as 1->2->3->4->5 with score 4. can you explain please.

[@anirbanghosh6328]

The solution link is not working

[@kaukuntlaprudviraj108]

it is failing at one test case